M1 June 2011 Q7 Vectors Question

I don't understand the logic for part d at all. I'd appreciate the help

[In this question, i and j are horizontal unit vectors due east and due north respectively
and position vectors are given with respect to a fixed origin.]
A ship S is moving along a straight line with constant velocity. At time t hours the position
vector of S is s km. When t = 0, s = 9i – 6j. When t = 4, s = 21i + 10j. Find
(a) the speed of S,

(b) the direction in which S is moving, giving your answer as a bearing.

(c) Show that s = (3t + 9) i + (4t – 6) j.

A lighthouse L is located at the point with position vector (18i + 6j) km. When t = T, the
ship S is 10 km from L.
(d) Find the possible values of T.

I don't understand the logic for part d at all. I'd appreciate the help

[In this question, i and j are horizontal unit vectors due east and due north respectively
and position vectors are given with respect to a fixed origin.]
A ship S is moving along a straight line with constant velocity. At time t hours the position
vector of S is s km. When t = 0, s = 9i – 6j. When t = 4, s = 21i + 10j. Find
(a) the speed of S,

(b) the direction in which S is moving, giving your answer as a bearing.

(c) Show that s = (3t + 9) i + (4t – 6) j.

A lighthouse L is located at the point with position vector (18i + 6j) km. When t = T, the
ship S is 10 km from L.
(d) Find the possible values of T.

I will write the vectors as column vectors as I think it is easier to see what is going on.
From part c we have

$\displaystyle S= \begin{pmatrix} 9 \\ -6 \end{pmatrix} +t \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
This is the vector equation of the line. We are trying to find when S is 10 away from the position vector
$\displaystyle \begin{pmatrix} 18 \\ 6 \end{pmatrix}$.
Notice how 10km is 2 times the magnitude of the direction vector of the path of S?
What can you deduce from that?

I will write the vectors as column vectors as I think it is easier to see what is going on.
From part c we have

$\displaystyle S= \begin{pmatrix} 9 \\ -6 \end{pmatrix} +t \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
This is the vector equation of the line. We are trying to find when S is 10 away from the position vector
$\displaystyle \begin{pmatrix} 18 \\ 6 \end{pmatrix}$.
Notice how 10km is 2 times the magnitude of the direction vector of the path of S?
What can you deduce from that?

That it's 2(3i + 4j)?


Posted from TSR Mobile